Guys, a dumb question: which one should I use: "parametrization" or "parametERization"? I rather use the first one, but it seems it is not grammatically correct? (I'm not a native English speaker)
are there any neat ways of constructing an injective but not surjective function $f: \mathbb{N}\to\mathbb{N}$ without using some arbitrary piecewise definition?
I wanted to just put it in a simple form $f(x) = ...$ without having to mess with brackets (idk how to do piecewise functions in latex)
suppose we take representations of $SU(2)$ on $S^2$. these do not contain the half integral reps what happens if we take the reps of SU(2) on $S^3$? Do these contain the half integral reps
I can't imagine $f(x) = \sqrt{x}$ or $\frac{1}{1-x}$ or something works, unless we say explicitly that this function is defined only for $f(x) \in \mathbb{N}$?
sorry, i meant reps of su(2) on $L^2(S^2)$and $L^2(S^3)$ respectively does the latter contain half integral reps
maybe the latter includes only half integral reps, like the former includes only integral ones? i am representing $su(2)$ on the latter using left invariant vector fields of SU(2) since this is faithful representation, does this only include half integral reps because the integral ones are not faithful, as it is many-to-one
Hi everyone. I was reading an answer on Physics SE where it is said: "$E(2v)=4E(v)$ [...] implies that $E$ is quadratic. Now, it doesn't seem obvious to me. How can we prove that $f(2x)=4f(x)$ doesn't have a non-quadratic solution?
As Wikipedia says:
[...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.
Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ ...
@Bml sorry but you'll have to do some translation work into mathematical language what's $E$ supposed to be? A function $E\colon \mathbb{R} \to \mathbb{R}$?
I am reading Folland's real analysis text, section 2.2 on integration of nonnegative functions. I am stuck at the definition of the integral of a simple function and how to show it is well-defined. There are related questions/answers to mine which I've included at the end together with a short co...
I posted a question on main. In Folland's text, he defines the integral of a simple function in terms of the unique standard representation. Then proves linearity, which is a bit involved. In Stein and Shakarchi's book, however, using basically the same definitions as in Folland, they first show the integral is well-defined and linearity follows easily from this.
I wonder, does the independence of representation follow from linearity? I don't see otherwise how to show it is well-defined given Folland's exposition.
@SineoftheTime I think "quadratic" means "dependent on the argument (velocity) squared".
@SineoftheTime Sorry, I don't understand. Why does the answerer say $E(m,v) = m E(v)$, and why it is corresponding to $E(m,v) = m E(1, v)$? Why does the answerer say $E(a v) = a^2 E(v)$, and why it is corresponding to $E(1, a v) = a^2 E(1, v)$?
@Bml hi. this notation is confusing. E can't simultaneously be a function of two variables on the left and of just one variable on the right. Use $g$ on the right like @SineoftheTime says
then for each equivalence class, we can define $E(x)$ for some arbitrary element of the class, and this fixes the value of $E(x)$ at all the other elements of the class
@BenSteffan yes..
@SineoftheTime from a physics point of view, u can assume an analytic function...and then only the quadratic term survives
@BenSteffan @SineoftheTime @RyderRude I went through the physics of this problem. Assuming $E(m,v) = m E(v)$, the most general case is $m E(v+u) + m E (v-u) = 2m E(\sqrt{v^2+u^2})$, i.e. $E(v+u) + E (v-u) = 2 E(\sqrt{v^2+u^2})$. What we examined is the case in which $u=v$, which reduces to $E(2v) = 2E(v \sqrt 2)$. I don't know how the answerer got $E(2v)=4E(v)$.
How to solve $E(v+u) + E (v-u) = 2 E(\sqrt{v^2+u^2})$
the problem is not solving it, since $x\mapsto x^2$ satisfy the functional equation. The problem it to show that is unique, and I'm not sure this is even correct without further hp
The games in the database in this line seem to be mostly recent. Every time I see a classical line but with an early h6/a6/h3/a3 thrown (or even h5 etc) in I assume it's an engine influenced idea
Because I know people really care about this: if I keep paying my mortgage the way I have been so far, it will be completely paid off in 2048, almost 3 years early! (yay?)
@BenSteffan Ok, let's try. $g(v) = g(v \cdot 1)$, and $g(v \cdot 1)$ is of the form $g(a \cdot k)$, with $k=1$ and $a=v$. So, since $g(av) = a^2 g(v)$, then $g(ak) = a^2 g(k)$, and setting $k=1$ and $a=v$ we have $g(v) = v^2 g(1)$. Setting $g(1) = c$ by definition, we have $g(v) = cv^2$. Are there other assumptions I'm missing?
As Wikipedia says:
[...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.
Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ ...
